A researcher is interested in determining the average number of years employees of a company stay with the company. If past information shows a standard deviation of 19 months, what size sample should be taken so that at 95% confidence, the margin of error will be 4 months or less?
z - z-score
e - margin of error
p - standard deviation
For 95% confidence z-score = 1.96
"\\\\n = \\frac{z^2\\times p^2}{e^2} = \\frac{1.96^2\\times19^2}{4^2}=" 87
Answer: 87.
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